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Exercise 2 a=c(21,36,42,24,25,36,35,200,32) mean(a) [1] tmean(a) [1] median(a) [1] 35 Sample mean is not resistant as its value largely inflated with a change of a single value.

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Exercise 3 The resistance of the 20% trimmed mean is 0.2, meaning that a change in more than 20% of the observations are required to generate a large change in its value. In this case n=9, so 9×0.2=1.8, rounded down to the nearest integer, =1. This means that a large change in the 20% trimmed mean value requires a change of at least 2 observations.

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Exercise 4 The resistance of the Median is 0.5, meaning that a change in more than 50% of the observations are required to generate a large change in its value. In this case n=9, so 9×0.5=4.5. This means that a large change in the median requires a change of at least 5 observations.

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Exercise 11 Yes, because we shift the extreme values closer to the mean. This reduces the dispersion in the data. The mean squared distances from the mean decreases accordingly.

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Exercise 12 The variance has s sample breakdown point of 1/n, thus a single observation can render it value arbitrarily large or small.

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Exercise 13 The sample breakdown point of the 20% Winsorized variance is 0.2. In the case of n=25, this would be 25×0.2= 5. Thus, we need at to change at least 6 observation to render the Winsorized variance arbitrarily large.

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Exercise 23 The boxplot has a sample break down point of 0.25%. The number of outliers it detects does not exceed 25% of the sample. For example, when we had 3 outliers with n=10, all outliers disappeared.